Op-Amp basic problem with a LED diode

[Femme_physics]

The positive supply is always connected at the top of the op-amp, regardless of where the + and - inputs are.

Ah, really? That's kinda funny. Well, in that case

Vout = 7.8 [V]

The negative supply is always connected at the bottom of the op-amp.

[qupte]What is connected at the bottom of the op-amp?[/quote]

Well, V+ (what I defined it as) is connected to the bottom of the op-amp. Therefor -Vout = V+ = -5.526

 

[I like Serena]

Ah, really? That's kinda funny. Well, in that case

Vout = 7.8 [V]

Yes. That is true in case 1.

Well, V+ (what I defined it as) is connected to the bottom of the op-amp. Therefor -Vout = V+ = -5.526

No, that is connected to the + input of the op-amp.
The negative power supply is specified to the right of that.

 

[Femme_physics]

I just realized something, I just need to find Vout in case 1 and case 2. Why did I do all those minuses and pluses for?

In both cases the Vout ways exceeds the limit of the op-amp. In fact, shouldn't the op-amp burn if that happens?

 

[I like Serena]

I just realized something, I just need to find Vout in case 1 and case 2. Why did I do all those minuses and pluses for?

Which minuses and pluses?

In both cases the Vout ways exceeds the limit of the op-amp. In fact, shouldn't the op-amp burn if that happens?

No, it just gets saturated and Vout is just either the positive or the negative power supply.

 

[Femme_physics]

Which minuses and pluses?

I noticed I was trying to find -Vout and +Vout in the last few posts which doesn't make sense, since there is only 1 Vout in each of the 2 circumstances. I don't know why I did that. Must've gotten confused.

No, it just gets saturated and Vout is just either the positive or the negative power supply.

Gotcha.

No, that is connected to the + input of the op-amp.
The negative power supply is specified to the right of that.

Vout2 = V- = 6.3V

?

 

[I like Serena]

Vout2 = V- = 6.3V

?

Nooooo... there is an Earth symbol there.
Shouldn't that be the negative power supply?

 

[Femme_physics]

Oh!

So, ok, taking the lower leg into consideration it's 7 - 5.526 = 1.474 [V]

 

[I like Serena]

Oh!

So, ok, taking the lower leg into consideration it's 7 - 5.526 = 1.474 [V]

Huh?

Where did you get 7 V?

 

[Femme_physics]

Well, that's V3

 

[I like Serena]

Well, that's V3

But... V3 is not connected to the op-amp.
Before you had already the proper V+ and V-...

 

[Femme_physics]

:(

I'm really confused and lost. A bit more clue-ins, please?

 

[gneill]

The op-amp has a pair of power connections. This is where the op-amp gets the power to do what it does. The op-amp can only provide an output voltage that lies between the positive power supply and the negative power supply voltages. If you apply inputs that would cause the op-amp to "want" to create an output that is higher than the positive supply voltage, or less than the negative supply voltage, the op-amp can only reach those limits.

What conditions will make the op-amp drive the output to one power supply level or the other? For an ideal op-amp, ANY voltage difference on the inputs will do it!

For this circuit the positive power supply for the op-amp is Vs = 7.8V. The negative supply voltage is zero because the negative supply lead for the op-amp is connected to ground (zero volts). So the output of the op amp is restricted to the range 0 ≤ Vo ≤ 7.8 .

Does that help?

 

[Femme_physics]

Are you telling me that Vout in case 2 = 0V ?

 

[gneill]

Are you telling me that Vout in case 2 = 0V ?

I suppose I am

 

[Femme_physics]

Ah...because no matter what's my V there, as long as it's in minus, it automatically turns into 0 because the transistor minus leg is ground?

EDIT:

BTW one of my classmates looked at me oddly when I presented this formula:

Vout = (V+ - V-) x 1000000

As though we don't need it to solve the problem. Is it needed, or not?

 

[gneill]

Ah...because no matter what's my V there, as long as it's in minus, it automatically turns into 0 because the transistor minus leg is ground?

Well, op-amp rather than transistor (an op-amp actually contains a whole bunch of transistors). The basic rules are: the op-amp output cannot exceed the power supply levels, and any voltage difference between its inputs will drive it to one level or the other.

This applies to ideal op-amps. Real ones come very close to this behavior, too. So much so that one can usually ignore any discrepancies for purposes of analysis.

EDIT: No, you don't really need that formula, but if it helps you to remember which direction the op-amp is going to drive the output by all means keep it handy. Maybe replace the 1000000 with a big G or A or something that won't attract undue remarks

 

[Femme_physics]

Well, op-amp rather than transistor (an op-amp actually contains a whole bunch of transistors).

Oops! Mistyped. meant op-amp :)

The basic rules are: the op-amp output cannot exceed the power supply levels, and any voltage difference between its inputs will drive it to one level or the other.

This applies to ideal op-amps. Real ones come very close to this behavior, too. So much so that one can usually ignore any discrepancies for purposes of analysis.

Ah, of course! I feel so stupid missing all the hints ILS gave me!

EDIT: No, you don't really need that formula, but if it helps you to remember which direction the op-amp is going to drive the output by all means keep it handy. Maybe replace the 1000000 with a big G or A or something that won't attract undue remarks

LOL. Got it

Thanks! Will scan full solution soon

 

[Femme_physics]

Does the way I present the solution looks okay?

http://img854.imageshack.us/img854/9623/thewayr.jpg

 

[I like Serena]

Looks fine to me!

 

[Femme_physics]

Terrific thanks!